Question

if alpha beta are zeros of the quadratic polynomial 2 x square - 4 x + 1 then Alpha square plus beta square is equals to​

Answer 1

Answer:

Alpa + beta = -b/a = 4/2=2

Alpa (beta) = c/a = 1/2

alpa²+ beta² = (alpa+ beta)² - 2(alpa)beta

= (2)²- 2(1/2)

= 4 - 2/2

= 8 - 2/2

alpa²+ beta² = 6/2= 3

Step-by-step explanation:

I hope it will help you!

Answer 2

Answer:

Required value of ${ \alpha }^{2} + { \beta }^{2}$ is 3

Step-by-step explanation:

Given

$\alpha \: and \: \beta$

are zeroes of the quadratic polynomial

$2 {x}^{2} - 4x + 1$

We know,

If $a {x}^{2} + bx + c$ is a polynomial and it has two zeroes then sum of two zeroes is $( - \frac{b}{a} )$ and product of two zeroes is $\frac{c}{a}$

Here

$\alpha \: and \: \beta$are two zeroes of given polynomial $2 {x}^{2} - 4x + 1$

So,

$\alpha + \beta = - ( - \frac{4}{2} ) \\ = 2$

and

$\alpha \beta = \frac{1}{2}$

Now,

${ \alpha }^{2} + { \beta }^{2} \\ = {( \alpha + \beta )}^{2} - 2 \alpha \beta \\ = {2}^{2} - 2 \times \frac{1}{2} \\ = 4 - 1 \\ = 3$

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549

#SPJ2